Finding Exact Value of Trig Expression

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whatisphysics
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Homework Statement



Find the exact value of each expression:
a) sec-1([tex]\sqrt{}2[/tex])
b) sin-1(1)

Homework Equations



sec[tex]\theta[/tex]=[tex]\stackrel{}{}1/cos\theta[/tex]

The Attempt at a Solution


I've never learned this, but I am really curious in how it is solved.
Is there a formula for this? Thanks!
 
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These are inverse functions so:

a. For what value of x does sec(x) = sqrt(2)

b. For what value of x does sin(x) = 1?
 
whatisphysics said:

Homework Statement



Find the exact value of each expression:
a) sec-1([tex]\sqrt{}2[/tex])
b) sin-1(1)

Homework Equations



sec[tex]\theta[/tex]=[tex]\stackrel{}{}1/cos\theta[/tex]

The Attempt at a Solution


I've never learned this, but I am really curious in how it is solved.
Is there a formula for this? Thanks!
Do you understand inverse functions?

IOW, x = f-1(y) <==> y = f(x)

For example, suppose you were asked to find cos-1(1/2).

Let y = cos-1(1/2).
That is equivalent to 1/2 = cos(y). What angle in the interval [0, [itex]\pi[/itex]] has a cosine of 1/2?
 
Mark44 said:
Do you understand inverse functions?

IOW, x = f-1(y) <==> y = f(x)

For example, suppose you were asked to find cos-1(1/2).

Let y = cos-1(1/2).
That is equivalent to 1/2 = cos(y). What angle in the interval [0, [itex]\pi[/itex]] has a cosine of 1/2?

Should I memorize the circle with all the angles?
And this may sound silly...but on (x,y), which is cos and sin? Is it like (cos, sin) on the circle, or the opposite?
 
A simple way to look at the problem is let [tex]a=\sec^{-1}\sqrt{2}[/tex] then [tex]\sec a=\sqrt{2}[/tex]. From here it is easy to compute the value a by turning sec into cos and using information about known values of cos.
 
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Thank you all for the input! I think I will learn to memorize the circle with all the angles...I'm sure that will help.